(Hope the Unicode math symbols appear correctly. )

> After the feather is released, the feather is gravityless (in the non inertial reference frame) so the barbules go to their natural position, so they go "up", but only slightly and the flutter is minimal.

You're saying the same thing as I did in different words, in fact my analysis was also in the accelerating frame :-).

So, in the accelerating frame at t=0 (the drop), we have an deformed elastic object that acts very similar to a harmonic oscillator, with some initial deformation Δx. It's not exactly a simple harmonic oscillator because the mass is distributed throughout, and the effective mass is NOT the mass of the spring (in fact it's one third of that), but it's relatively easy to show it moves like one (to the first degree for small displacements), albeit not in a Hacker News discussion. It will move according to

    D²[x] + 2ζωD[x] + ω²x = 0

where ζ=c/2√(mk) is the damping ratio, ω=√(k/m) is the undamped angular frequency and m is the effective mass. For a feather, it might well be that the oscillator is overdamped, of course. The ball most certainly has a much larger Q factor than the feather.

Elasticity and sound waves are very closely related. Sound is a particular kind of wave traveling through an elastic medium. For an approximately linear object, we only have longitudinal waves that move with c=√(Y/ρ). Y is Young's modulus and ρ is the density. You can immediately see that small values of Y give rise to small values of c.

For ideal springs of length X and cross-section S we can prove that (exercise left to the reader) that:

    Y = Xk/S

so

    kΔx/S = Y Δx/X => F/S = Y Δx/X => Y = FX/SΔx

I can estimate X=0.2m and S=1E-5 (maybe larger than you expected because we need to consider the whole feather as the spring, not just its stem (wrong term?). Under it's own weight the feather seems to move approximately Δx=1E-2m so X/Δx=20, F is of course half the weight. It's hard to get good estimates on the internet for the weight of a feather. The most reliable estimates seem to be 8.2E-6kg for a chicken feather. Chicken feathers are small, let's say 5E-2m so our feathers have a mass of 8.2E-6kg·(X/5E-2m)^3=0.5E-3kg. Half a gram for the giant feather. Doesn't sound implausible.

Plugging all values in we get Y=5248Pa. Speed of sound is c=√(Y/ρ)=c=√(Y/(M/V))=√(YSX/M)=4.58m/s, a very low value indeed and quite consistent with the slow motion video.

Just for kicks, the speed of sound in rubber 60m/s. It shouldn't surprise us at all that the speed in feathers is smaller than for rubber. Feathers are much weaker springs than usual rubber springs. (Note that if we had a spring made of rubber coils, the speed of sound in the spring would also be much smaller than the speed of sound in a rubber rod).

The approximations made above are all pretty wild, but the point I was trying to make is that waves travel quite slow in weak springs. The exact values don't matter. Just get yourself a long spring made of plastic or thin metal wire and create some waves. You can see the propagations, they move very slowly.

I think that your model is correct, but I think that it's possible to add a few simplifications.

If the speed of sound is ~5m/s and the feather length is 0,2m then the time to for a signal to travel from one tip to the other is ~1/25s. That is approximately one frame in a standard camera (PAL/NTSC). So if they are using a standard camera, it's possible to assume that the drop signal travels almost instantly.

The system is totally overdamped, so I think that it's better to model it as a movement in a fluid with hight viscosity (like a spring inside honey). If you drop the acceleration term and you get

2ζω.D[x] + ω²x = 0

where 2ζω/ρ is something like the apparent viscosity that see a barbule moving inside the "fluid" of the other barbules.

The solution is

X = A exp(-ω/2ζ t)

So the relaxation time is ω/2ζ. When it's overdamped the system almost don't oscillates. Watching at the video, after it falls 1-ball-diameter the barbules reach the final position. I guess the ball is ~0,3m, so I guess that ω/2ζ ~= Sqrt(2h/g) = 0,8 seconds ~= 20 frames in a standard camera. But I have not idea about a sensible value for ζ.